Wikipedia:Reference desk/Archives/Science/2022 November 30

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November 30

In Rutherford’s gold foil experiment, why did Rutherford use an alpha particle a billion times larger than a gold atom?

An alpha particle has root-mean-square charge radius of 1.67824(83) femtometers. The volume of a gold atom is 1.25 x 10^-23 cm3. The gold foil thickness was 100 atoms. Therefore, multiple gold molecules deflect every alpha particle. The experiment does not definitively indicate that a molecule has a nucleus. Vze2wgsm1 (talk) 13:18, 30 November 2022 (UTC)

Your objection is over a hundred years too late. Geiger–Marsden experiments describes the actual experiment, if you can produce the maths giving a good approximation to the experimental model using what they called the plum pudding model of the atom or some other model without a small nucleus I'm sure it would be of interest.. NadVolum (talk) 14:04, 30 November 2022 (UTC)

The alpha particle and gold never touch each other.

$\Delta p=F\Delta t=k\cdot {\frac {Q_{\alpha }Q_{n}}{r^{2}}}\cdot {\frac {2r}{v_{\alpha }}}$

If a high-speed electron headed directly toward another electron, each would feel the same repulsive force. Over time, both would receive equal KE, due to the force. Therefore, at best, the fast electron would lose half its speed and the slow electron would gain an equal amount of speed. The stationary electron could reverse the direction of the fast electron by remaining stationary. Did Rutherford assume the Gold nucleus remained stationary? Vze2wgsm1 (talk) 19:44, 6 December 2022 (UTC)

(edit conflict)Eh? From Geiger–Marsden experiments (which describes and explains the experiment) "radius of a gold atom = 1.44×10−10 m", whilst from Helium-4 "The size of the 4He nucleus has long been known to be in the order of magnitude of 1 fm". Now 1 fm is 10^-15m or 100,000 times smaller than a gold atom. Intuitively, a gold nucleus contains 79 particles, an alpha particle contains 4, so you would expect the alpha to be smaller than the gold's nucleus, let alone the atomic size. Martin of Sheffield (talk) 14:08, 30 November 2022 (UTC)

I'm going to take a flyer here and guess that Vze2wgsm1 has just made a very simple arithmetic mistake, or perhaps a mistake of dimensional analysis. He/she quotes the volume of the gold atom as 1.25 × 10⁻²³ cm³, which would be 1.25 × 10⁻²⁹ m³, which is somewhere in the ballpark for the volume of a sphere of radius around 10⁻¹⁰ m. However he/she then seems to go on to compare this 10⁻²³ number to the 10⁻¹⁵ m order of magnitude for the linear dimension of the alpha particle. But those two orders of magnitude are in completely different units, and comparing them in that way simply doesn't make sense. If you cube the linear dimension to get the volume, you still get a quick-and-dirty answer, but one that's more indicative of what's going on. --Trovatore (talk) 08:27, 1 December 2022 (UTC)

Thanks Trovatore. I apologize for the sloppy math. Vze2wgsm1 (talk) 19:55, 6 December 2022 (UTC)

Thanks (Martin of Sheffield). Size counts in collisions. Rutherford's calculations are based on force between point charges, without collisions. Vze2wgsm1 (talk) 19:53, 6 December 2022 (UTC)

It should be noted that Rutherford et al developed mathematical models to match the data they got from the experiment. The mathematical models that fit the data best were small point-like particles with very high mass, arranged at a significant distance from each other. Unless you have a mathematical model that fits the data better, then you have no leg to stand on to criticize the model that Rutherford et all developed to explain the results of their experiment. The Wikipedia article itself on the experiment only touches briefly on these models, and provides a lay-accessible explanation of the conclusions drawn from the modeling so done. If you want to know all of the details (and you need to know all of the details if you're going to decide for yourself if it is wrong or right) then you need to go to the source papers. They're publicly available. The math isn't hard, per se, but it is involved, and you're going to need to take some time to get a grasp of it. --Jayron32 15:36, 30 November 2022 (UTC)

I'm inclined to ask - how does anyone know the size of either particle - ... ?

Isn't the size of the particle based on theory that is built upon experiment, and if we're being specific, isn't the experiment one that is structurally similar to the foundational scattering experiment?

In other words, we know the size of the particles because that's the size that is most consistent with the bouncing-off-ish behavior we observe.

The century-old version of this bouncing-off-experiment has been substantially improved, and one of the things that makes a university-physics-program interesting is that you get to take a lab class and do it yourself!

Here's Rutherford Scattering from MIT's OpenCourseWare program - "quite similar to that in the Junior Lab" (that's one of the classes that the third-year physics students get to participate in!)

If there were something substantially invalid about the theory - if the radii were off by orders of magnitude, or if the basic premise of the experiment was not a suitable model - that's the sort of thing that lots of people would have noticed, because lots of people reproduce the original experiment or its modern variations. So - argumentum ad populum notwithstanding - and with great awareness about the history of experimental physics (we have a Wikipedia article on that, too)... the long and short of it is that you probably aren't the first person to question the validity of a very old physics experiment, so if it's important to you, it's worth spending time to investigate the details of those questions that have already been answered.

Nimur (talk) 21:36, 30 November 2022 (UTC)

Thanks.

Why did every alpha particle have some deflection? The cross section of a gold atom is 3.45 x 10^8 x the cross section area of gold’s nucleus. If an alpha particle passed through the cross section of a gold atom, its chance of any deflection by the nucleus would be one in 3.45 x 10^8 Vze2wgsm1 (talk) 08:41, 1 December 2022 (UTC)

Alpha particles can also be deflected by the electrical field of the electron cloud. In the actual experiment, the vast majority of the incident alpha particles had only very small deflections,^[1] which means that they never got close to a nucleus.  --Lambiam 10:45, 1 December 2022 (UTC)

Deflection between point charges would not necessarily indicate their radius, due to potentially infinite force between charges. Vze2wgsm1 (talk) 12:25, 1 December 2022 (UTC)

Charges obey the inverse square law, something that was known a long time before Rutherford, see Coulomb's law, a law developed well over a century before Rutherford was working, and well tested and established by that time. A charged surface of a meaningful particle size (something with a sizable volume) will behave differently than a point particle will. Rutherford's data showed that treating both the alpha particle and the gold nucleus as point particles produced the best results. Note that this doesn't mean that the particles in question were point particles strictly speaking, just that they were smaller than the significant figures of his calculations allowed him to see; basically the resolution of his experiments did not allow for him to resolve the size of the nuclei, and therefore they were quite small indeed. --Jayron32 14:00, 1 December 2022 (UTC)

Thanks Jayron. The Rutherford model should not be considered gospel unless the Rutherford model is better than the other published models. Rutherford's model assumed point charge repulsion (without collision) caused alpha particle recoil. Is Rutherford's model better than a model that assumes alpha particle collisions with gold molecules containing various numbers of gold atoms? Vze2wgsm1 (talk) 20:29, 6 December 2022 (UTC)

The short answer is yes, the Rutherford model is a far better model and predicter the observed scattering behavior than the plum pudding model or "a model that assumes alpha particle collisions with gold molecules containing various numbers of gold atoms." The long answer is that nothing, nothing in science is treated as "gospel." The Rutherford model was better than the models that came before it, but it has been refined and succeeded by better models. Remember that you are discussing a model here from over 110 years ago. Since then, we have the Bohr model, which has itself been succeeded by other more modern quantum mechanical models, such as the valence shell model. None of this is gospel, they are just better and better descriptions of what we experimentally observe. We've done a lot more experiments in the last 110 years; the last experiment and best experiment was not the gold foil experiment. However, it has a valued place in scientific history as it put us on the track to far better and more accurate models of the atom than we had before it. The difference between plum pudding and the Rutherford model is a great one, while the difference between the Rutherford model and more modern models is more in the details (especially the math). --OuroborosCobra (talk) 21:23, 6 December 2022 (UTC)

I think Einstein was the first to determine the size of atoms with his explanation of Brownian motion. NadVolum (talk) 15:04, 1 December 2022 (UTC)

Actually, credit is usually given to others a bit later. Einstein's Brownian Motion paper is usually credited with proving the physical existence of atoms as real things (rather than merely convenient models), but it was the Van der Waals equations of state that was the first to establish that atoms had real physical volumes that could be estimated by deviations from the ideal gas law, while Bragg's law allowed for the determination of the sizes of atoms in solid crystal lattices to surprising accuracy. It should be noted that ALL of these things were happening pretty much simultaneously in the period from about 1908-1913 or so. It wasn't like one was waiting on the results of the other; each of these labs (and MANY MANY more) were all approaching the analysis of atomic structure from a variety of perspectives and experimental techniques. --Jayron32 16:13, 1 December 2022 (UTC)

(I'm a bit late - but I endorse what Jayron wrote). And, though I ramble, let me try to summarize: "no," one scientist was not the scientist who determined the size of the atomic nucleus. It turns out to be a bit more complicated. Hey, it's particle physics, what did you expect?

I think that "the size of atoms" is a conceptual focal-point in the study of physics; and it's an area of inquiry that has evolved over many centuries, increasingly informed by modern scientific experiment and modern mathematical treatments; so if we try to credit this to one individual, or one experiment, that's an oversimplification - a reductio ad absurdum, (I'm on a logical fallacy kick this week!)

Many famous scientists, and many more non-famous scientists, have contributed to the ongoing knowledge and theory that helps us to conceptualize (and even just to define) the "size" of particles that constitute our universe. Perhaps a gloss over our article on the history of subatomic physics, or a historical review from the Plato Encyclopedia of Philosophy ? During my early education, I studied something called the theory of knowledge - formally studying that topic can help us to dispel ideas about "single individuals" who "discover" "science" - because when we perpetuate sound-bites like "Einstein discovered Brownian Motion," this is not only factually incorrect in its detail, it's also categorically destructive, in that the perpetuation of the concept elevates invalid constructs, reinforces harmful social structures, and diminishes the importance of other contributors. Individual facts are hypothesized in individual experiments; these facts are established and accepted by repeated experiment; facts are distilled into theory over the course of time; but knowledge is something that permeates a community. "We know how large a particle is" - this statement is so troublesome! Who are we? (Which community knows this, and what percentage agree on every detail?) What do we know? If these questions were even half as easy as the particle-physics equations that we initially seemed to be considering, there'd be a lot more particle-physics knowledge, dispersed throughout the whole the world, and not compartmentalized inside the incomprehensible technobabble of academic publications that are known only to a tiny number of humans.

One of the most key insights is that the "size of a particle" depends on context - and the context often depends on the community of people who are using this factoid for some purpose, practical or otherwise. There are many different sizes - each parameter is relevant to a different physical process, and manifests in a different way in various experiments. I'm a fan of the collision cross-section measurement, because that parameter is useful to me - but in my mental model, this "useful parameter" is actually not a great conceptual tool for knowing the "size of the thing," whether it's an alpha particle or a gold nucleus or an F-35 jet... Sometimes there is a scale-invariant size that incontrovertibly describes a thing; and other times, the size changes depending on how you look at it!

Nimur (talk) 16:29, 1 December 2022 (UTC)

Excellent summary of the Epistemology of scientific discovery; as always, a lot is said, but a lot needs to be said, to capture these fundamental problems of knowledge building. One thing I also want to add to this is the fundamental problem with trying to describe atomic-scale physics by analogy to human-scale physics. We tend to thing of concepts like "distance" and "size" and "object" as so self-evident and fundamental that we treat them as axiomatic: they just are and don't need analysis or probing. I mean, I know what an "thing" is self-evidently. Like I can hold a billiard ball in my hand. It's a thing; it has a surface and an inside and I can touch the outside and measure it, and positively identify where it is and where it isn't all without any real measuring device more complex that my own senses. So it is natural to think that things like "atoms" are basically like that, but just smaller. Except it isn't like that at all. Even self-evident concepts like what is an object or a thing basically break down to the point where it has no meaning, or a meaning so different from our experiences that using analogy only leads us astray. Are atoms spherical objects? Yes, kinda-sorta, but they aren't like billiard balls, and they aren't even like clay balls (you know, maybe they're spherical and squishy? that could explain the different kinds of atomic radius? Nope, wrong again). Literally any analogy you could make to try to understand an atom by connecting it to the behavior of anything you have experienced with your senses is so wrong it will just lead you to make false predictions about how atoms behave. We know how atoms behave because we have mathematical models that are, quite literally, the most accurate physical models ever created. The behavior of atoms and subatomic particles can be predicted and observed to match predictions using these models to within an accuracy greater than a part per billion or more. Quantum mechanics has been described as the most accurate physical theory ever developed. It works so well. And yet, it tells us nothing more than a bunch of really hard-to-solve mathematics equations that we plug numbers into, get numbers out of, and then make measurements of atoms and subatomic particles to check to see how well it matches. What do those numbers tell us about what atoms look like? Fucked if I know. I'm not sure anyone does. Abandoning the need to make visually coherent analogies to our own sensual experiences is the only way to make QM make sense. It works. Shut up and calculate. --Jayron32 17:33, 1 December 2022 (UTC)

The claim that we cannot explain atomic physics may be going too far. More accurately, we are searching for an explanation of atomic physics.

A lot of scenarios can comply with some of our observations of atomic physics, but not all. The Sagnac effect could be explained in the GPS system, if the Earth's atoms became geostationary during light travel. Light speed would be relative to geostationary while atom speed is relative to geostationary. However, all the Earth's atoms becoming geostationary during every quantized molecular energy change is not possible.

Further thoughts, based on the impossible: While we are geostationary, our clocks would not be ticking. Returning relativistic KE to molecules could require additional geostationary time. Average particle speed relative to geostationary could be less than actual speed. The additional geostationary time could explain time dilation. Vze2wgsm1 (talk) 12:05, 2 December 2022 (UTC)

I never once claimed we cannot explain atomic physics. I said we cannot explain atomic physics using analogy to object behavior in the macroscopic world. Atomic physics is very well explained by quantum mechanics. You need to use the proper explanations. --Jayron32 15:07, 2 December 2022 (UTC)

Molecules are supposed to be a container of atoms. Molecules are containers of non-quantized energies. Large molecules can exist during a quantized molecular energy change.

If a single H2O unit supplied the enthalpy of evaporation for another H2O unit, the heat source H2O must be 538 ^oC hotter than the evaporating H2O.

An existing H2O unit does not increase volume 1700x during evaporation during zero time.  Phase changes occur without inertia penalty.

If a 100 ^oC steam H2O unit within air deposits its enthalpy of evaporation into a single (N2) molecule the N2 molecule must be 1962 ^oC colder than the H2O unit.  Phase changes are constant temperature, so the adjacent molecule is the same temperature. 

Consider that light travels in a straight line through nuclear and molecular boundaries. If voids or high matter concentration existed, they must refract light, because matter slows light.

No refraction means homogenous material, no small molecules, no molecule movement, no relativistic kinetic energy.

If the entire Earth were composed of homogenous material, then the Earth would be geostationary, and the International Atomic clock would not be ticking. Vze2wgsm1 (talk) 16:39, 2 December 2022 (UTC)

Look, I'm not sure what to say now. Your statements are becoming increasingly incoherent, bordering on word salad. I can't even make out a coherent thought in the above rambling to even call it a misconception that could be corrected. It's a series of mostly unconnected statements of with a bunch of chemistry and physics related words thrown together into some bizarre stew. You're going to have to restate your query into something that makes at least some kind of sense if anyone is going to be able to converse with you. --Jayron32 17:33, 2 December 2022 (UTC)

The equation for the temperature change required for a single H₂O to supply enthalpy of vaporization to and evaporating H₂ is:

$dT={\frac {40613J}{75.38J/moloC}}=538$ ^oC

The evaporation takes place during a single quantized molecular energy change. The evaporating molecule is surrounded by same-temperature H₂O molecules. Therefore, every atom that supplies heat to the evaporating molecule must be included within the evaporating molecule.

Which makes my point that macroscopic molecules normally exist during phase changes.

An electrical circuit is a single molecule container of electrostatic and magnetic energy. Otherwise, the oscillation of an LC circuit would not be a function of total capacitance and total inductance. Which makes my point that molecules are containers of non-quantized energies.

Point charge theory (an atom containing a nucleus) does not explain why within an LC circuit, conversions of electrostatic to/from magnetic energy can proceed until exhaustion, instead of equilibrium. While magnetic energy is more stable than electrostatic energy, electrostatic energy can exhaustively convert into magnetic energy.

The conversions until exhaustion occur because the circuit’s ‘molecular’ energy cannot simultaneously be optimized for both electrostatic and magnetic energy. While magnetic energy is more stable than electrostatic energy, electrostatic energy can exhaustively convert into magnetic energy.

After exhaustion, the circuit still needs both electrostatic and/or magnetic energies for overall stability. A chemical change can cause the recently exhausted energy to become the more stable energy. Vze2wgsm1 (talk) 12:46, 4 December 2022 (UTC)

Brownian motion has perfect elasticity. An alpha particle hitting a gold nucleus does not.

In the world of imperfect elasticity, in billiards, a set of billiard balls can store the energy of a slowly moving cue ball as compression energy. The compression energy can return to the cue ball as KE, in the opposite direction. The rack cannot store the KE of a cue ball traveling at light speeds. No bounce. What do you think? Vze2wgsm1 (talk) 11:31, 2 December 2022 (UTC)

Again, I don't know how to say this any better, so I'll make it simple. You're just wrong. Atoms are not billiard balls, and they don't behave like that. Any analogy you make between the behavior of a billiard ball and the behavior of an atom (or any part of an atom like an electron or a nucleus) is going to lead you to wrong conclusions about how atoms behave. You can't just pretend like quantum mechanics is not a thing. --Jayron32 12:58, 2 December 2022 (UTC)

It's certainly true in general that thinking of atoms or nuclei as billiard balls can lead to incorrect conclusions because it ignores quantum effects, but I don't think that's what happened here. It looks like Vze2wgsm1 made a much more elementary mistake. See my first response. --Trovatore (talk) 17:41, 2 December 2022 (UTC)

While that is true for the initial question, further questions and statements from said user required additional misconception correcting. Also, while I appreciate your endorsement, the phrasing "ignores quantum effects" (which you didn't invent, it's a common thing to say) is something that gives me pause. When someone says that, it sounds like "quantum effects" are a kind of correction made to standard "classical physics" to fix some small deviation in the equations, or that it is something "tacked on" to standard physics. Not that you in particular believe this, but that kind of language does perpetuate the misconceptions I was trying to steer off. Atomic behavior obeys quantum mechanics full stop; the mathematics of QM is a coherent and sufficient physics to fully explain atomic behavior, and it isn't an "effect" per say. It's not the ignoring of some small correction, it's the ignoring of the entirety of the way atomic behavior works. --Jayron32 18:18, 2 December 2022 (UTC)

At a foundational level that's true, but it's not very practical. You can't just tell people to solve Schrödinger's equation for a Hamiltonian with an enormous number of variables every time they want to understand something. Even if you just want to dump it into a computer and come up with a numerical solution, you don't necessarily get much insight.

In practice, anyone who uses quantum mechanics for anything more complicated than a hydrogen atom needs to learn various approximations (mean-field approximation, geometric optics, etc etc), and often these take the form of a classical or semi-classical description, possibly combined with an extra layer of "quantum effects". In many situations that extra layer can be dropped without much harm.

In the specific case of the Rutherford experiment, as far as I can tell a classical billiard-ball description works just fine for explaining qualitatively why it was surprising, and it's quantitative enough that the OP's initial calculation would have been relevant if it had been done correctly. As for the further statements, I haven't tried very hard to read them — the first one was hard enough work already. --Trovatore (talk) 18:37, 2 December 2022 (UTC)

"At a foundational level that's true, but it's not very practical. You can't just tell people to solve Schrödinger's equation for a Hamiltonian with an enormous number of variables every time they want to understand something. Even if you just want to dump it into a computer and come up with a numerical solution, you don't necessarily get much insight." That's true, but if you can't solve the material yourself, you at least have to be able to trust the people that did when they tell you what the right answer is. It's fine to not have the skills or training to derive everything from first principles yourself, but when someone who can do that tells you what atoms really behave like, you have no footing to argue against that. It's fine to not know everything, but what it isn't fine to do is to argue from a place of ignorance against people that do know what they are doing. For the purpose of the OP, they appear to be arguing with Rutherford et al (along with the thousands of others who worked with, or came after Rutherford) that they were wrong. That's some gumption, especially for someone who doesn't have the background to argue from. Heck, I don't have the background to argue from. But I also recognize that, so I trust people that do. --Jayron32 19:18, 2 December 2022 (UTC)